Codeforces Round #273 (Div. 2)-A. Initial Bet
http://codeforces.com/contest/478/problem/A
1 second
256 megabytes
standard input
standard output
There are five people playing a game called "Generosity". Each person gives some non-zero number of coins b as an initial bet. After all players make their bets of b coins, the following operation is repeated for several times: a coin is passed from one player to some other player.
Your task is to write a program that can, given the number of coins each player has at the end of the game, determine the size b of the initial bet or find out that such outcome of the game cannot be obtained for any positive number of coins b in the initial bet.
The input consists of a single line containing five integers c1, c2, c3, c4 and c5 — the number of coins that the first, second, third, fourth and fifth players respectively have at the end of the game (0 ≤ c1, c2, c3, c4, c5 ≤ 100).
Print the only line containing a single positive integer b — the number of coins in the initial bet of each player. If there is no such value of b, then print the only value "-1" (quotes for clarity).
2 5 4 0 4
3
4 5 9 2 1
-1
In the first sample the following sequence of operations is possible:
- One coin is passed from the fourth player to the second player;
- One coin is passed from the fourth player to the fifth player;
- One coin is passed from the first player to the third player;
- One coin is passed from the fourth player to the second player.
解题思路:能够将硬币平均分给5个人,有一个trick就是开始都为0
1 #include <stdio.h>
2
3 int main(){
4 int a, sum = , t = ;
5 while(t--){
6 scanf("%d", &a);
7 sum += a;
8 }
9 printf("%d\n", sum >= && sum % == ? sum / : -);
return ;
}
Codeforces Round #273 (Div. 2)-A. Initial Bet的更多相关文章
- 贪心 Codeforces Round #273 (Div. 2) C. Table Decorations
题目传送门 /* 贪心:排序后,当a[3] > 2 * (a[1] + a[2]), 可以最多的2个,其他的都是1个,ggr,ggb, ggr... ans = a[1] + a[2]; 或先2 ...
- Codeforces Round #273 (Div. 2)
A. Initial Bet 题意:给出5个数,判断它们的和是否为5的倍数,注意和为0的情况 #include<iostream> #include<cstdio> #incl ...
- Codeforces Round #273 (Div. 2) A , B , C 水,数学,贪心
A. Initial Bet time limit per test 1 second memory limit per test 256 megabytes input standard input ...
- Codeforces Round #273 (Div. 2)-C. Table Decorations
http://codeforces.com/contest/478/problem/C C. Table Decorations time limit per test 1 second memory ...
- Codeforces Round #273 (Div. 2)-B. Random Teams
http://codeforces.com/contest/478/problem/B B. Random Teams time limit per test 1 second memory limi ...
- Codeforces Round #273 (Div. 2) D. Red-Green Towers 背包dp
D. Red-Green Towers time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- codeforces Codeforces Round #273 (Div. 2) 478B
B. Random Teams time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
- codeforces 的 Codeforces Round #273 (Div. 2) --C Table Decorations
C. Table Decorations time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Codeforces Round #273 (Div. 2)D. Red-Green Towers DP
D. Red-Green Towers There are r red and g green blocks for construction of the red-green tower. Re ...
随机推荐
- 51nod1010【二分】
打表+二分 #include <bits/stdc++.h> using namespace std; typedef long long LL; const LL inf=1e18+10 ...
- 一个高效的A-star寻路算法(八方向)(
这种写法比较垃圾,表现在每次搜索一个点要遍历整个地图那么大的数组,如果地图为256*256,每次搜索都要执行65535次,如果遍历多个点就是n*65535,速度上实在是太垃圾了 简单说下思路,以后补充 ...
- [Xcode 实际操作]九、实用进阶-(3)给代码方法添加宏注释
目录:[Swift]Xcode实际操作 本文将演示如何在方法列表中,对方法名称进行注释. 这样可以使程序,按功能分块,使方法清晰.易读并且方便定位. 在项目导航区,打开视图控制器的代码文件[ViewC ...
- route(2018.10.24)
建出最短路图之后\(topsort\)即可. 具体思路: 先用\(dijkstra\)算法在原图中跑出\(1\)号点到\(i\)号节点的最短距离\(dist_1(i)\),将所有边反向后用\(dijk ...
- 给mysql默认root用户设置密码
情况1:如果忘记了 MySQL 的 root 密码,可以用以下方法重新设置: 情况2:添加了phpmyadmin,登录时使用mysql的用户密码登录,但phpmyadmin不能使用空密码登录,即为 ...
- PAT甲级——1126 Eulerian Path
我是先在CSDN上发布的这篇文章:https://blog.csdn.net/weixin_44385565/article/details/89155050 1126 Eulerian Path ( ...
- python 基础(七) 异常处理
异常处理 一.需求 当遇到错误的时候 不让程序停止执行 而是越过错误继续执行 二.主体结构 (抓取所有异常) try: 可能出现异常的代码段 except: 出现异常以后的处理 三.处理特 ...
- 解决 mac zsh 所有命令失效
上面的没啥用, 直接看分割线吧, 上面的是第一次遇到这个问题, 没有解决.. zsh: command not found: 参考: https://www.jiloc.com/43492.html ...
- SSIS Passing Parameters to an ADO .NET Source query;向ado.net数据源传递参数。
使用SSIS的oledb数据源时的参数按钮如下图: 但是在使用ADO.NET源连接到MYSQL时,没有这个参数按钮,如何向数据流的sql command传递参数呢? steps: 1. 在 控制流 选 ...
- 23 在java中使用groovy类
1 在java中使用groovy类 1.1 直接调用groovy类 在java中调用Groovy类,需要增加Groovy运行时到java的classpath中. pom.xml < ...